Let A1C1 meet A2C2 at M'. Using Law of sine, M'C2=sin(<CC1A1)*C1C2/sin(<C1M'C2)and M'A2=sin(<A2A1M')*A1A2/sin(<A1M'A2)Evaluate M'C2/M'A2=(sin(<CC1A1)*C1C2)/(sin(<A2A1M')*A1A2),but sin(<CC1A1)/sin(<A2A1M')=A1C/AC1, which simplifies M'C2/M'A2 to (A1C*C1C2)/(AC1*A1A2)=A1C/A1A2*C1C2/AC1=1 because triangles AC1C2 and CA1A2 are similar.This proves M' to be midpoint of A2C2